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Ta có
$\dfrac{\arctan x}{x^2}=-\dfrac{ x}{x^2+1}+\dfrac{ 1}{x}-\dfrac{\dfrac{ x}{x^2+1}-\arctan x}{x^2}$
$\dfrac{\arctan x}{x^2}=\left (-\dfrac{ 1}{2}\ln(x^2+1) \right )'+\left (\ln x \right )'-\left (\dfrac{\arctan x}{x} \right )'$
$\dfrac{\arctan x}{x^2}=\left (\ln\dfrac{ x}{\sqrt{x^2+1}} \right )'-\left (\dfrac{\arctan x}{x} \right )'$
Suy ra
$\int\limits_{1}^{+\infty }\dfrac{\arctan x}{x^2}dx=\left[ {\ln\dfrac{ x}{\sqrt{x^2+1}}-\dfrac{\arctan x}{x}} \right]_{1}^{+\infty }=\dfrac{\pi+\ln 4}{4}$
Chú ý rằng
$\lim_{x \to +\infty}\ln\dfrac{ x}{\sqrt{x^2+1}} =\ln\left ( \lim_{x \to +\infty}\dfrac{ x}{\sqrt{x^2+1}} \right )=\ln 1 =0$
$\lim_{x \to +\infty}\dfrac{\arctan x}{x} \underbrace{=}_{L'Hospital}\lim_{x \to +\infty}\dfrac{(\arctan x)'}{(x)'}=\lim_{x \to +\infty}\dfrac{\dfrac{1}{x^2+1}}{1} =0$
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